(
Name
Gun_time
Net_time
Pace
John
28:48:00
28:47:00
4:38:00
George
29:11:00
29:10:00
4:42:00
Mike
29:38:00
29:37:00
4:46:00
Sarah
29:46:00
29:46:00
4:48:00
Roy
30:31:00
30:30:00
4:55:00
Q1. How can I add another column stating difference between Gun_time and Net_time?
Q2. How will I calculate the mean for Gun_time and Net_time. Please help!
I have tried doing the following but it doesn't work
df['Difference'] = df['Gun_time'] - df['Net_time']
for mean value I tried df['Gun_time'].mean
but it doesn't work either, please help!
Q.3 What if we have times in 28:48 (minutes and seconds) format and not 28:48:00 the function gives out a value error.
ValueError: expected hh:mm:ss format
Convert your columns to dtype timedelta, e.g. like
for col in ("Gun_time", "Net_time", "Pace"):
df[col] = pd.to_timedelta(df[col])
Now you can do calculations like
df['Gun_time'].mean()
# Timedelta('1 days 05:34:48')
or
df['Difference'] = df['Gun_time'] - df['Net_time']
#df['Difference']
# 0 0 days 00:01:00
# 1 0 days 00:01:00
# 2 0 days 00:01:00
# 3 0 days 00:00:00
# 4 0 days 00:01:00
# Name: Difference, dtype: timedelta64[ns]
If you need nicer output to string, you can use
def timedeltaToString(td):
hours, remainder = divmod(td.total_seconds(), 3600)
minutes, seconds = divmod(remainder, 60)
return f"{int(hours):02d}:{int(minutes):02d}:{int(seconds):02d}"
df['diffString'] = df['Difference'].apply(timedeltaToString)
# df['diffString']
# 0 00:01:00
# 1 00:01:00
# 2 00:01:00
# 3 00:00:00
# 4 00:01:00
#Name: diffString, dtype: object
See also Format timedelta to string.
I have the following column
Time
2:00
00:13
1:00
00:24
in object format (strings). This time refers to hours and minutes ago from a time that I need to use as a start: 8:00 (it might change; in this example is 8:00).
Since the times in the column Time are referring to hours/minutes ago, what I would like to expect should be
Time
6:00
07:47
7:00
07:36
calculated as time difference (e.g. 8:00 - 2:00).
However, I am having difficulties in doing this calculation and transform the result in a datetime (keeping only hours and minutes).
I hope you can help me.
Since the Time columns contains only Hour:Minute I suggest using timedelta instead of datetime:
df['Time'] = pd.to_timedelta(df.Time+':00')
df['Start_Time'] = pd.to_timedelta('8:00:00') - df['Time']
Output:
Time Start_Time
0 02:00:00 06:00:00
1 00:13:00 07:47:00
2 01:00:00 07:00:00
3 00:24:00 07:36:00
you can do it using pd.to_datetime.
ref = pd.to_datetime('08:00') #here define the hour of reference
s = ref-pd.to_datetime(df['Time'])
print (s)
0 06:00:00
1 07:47:00
2 07:00:00
3 07:36:00
Name: Time, dtype: timedelta64[ns]
This return a series, that can be change to a dataframe with s.to_frame() for example
My dataset looks like this:
time Open
2017-01-01 00:00:00 1.219690
2017-01-01 01:00:00 1.688490
2017-01-01 02:00:00 1.015285
2017-01-01 03:00:00 1.357672
2017-01-01 04:00:00 1.293786
2017-01-01 05:00:00 1.040048
2017-01-01 06:00:00 1.225080
2017-01-01 07:00:00 1.145402
...., ....
2017-12-31 23:00:00 1.145402
I want to find the sum between the time-range specified and save it to new dataframe.
let's say,
I want to find the sum between 2017-01-01 22:00:00 and 2017-01-02 04:00:00. This is the sum of 6 hours between 2 days. I want to find the sum of the data in the time-range such as 10 PM to next day 4 AM and put it in a different data frame for example df_timerange_sum. Please note that we are doing sum of time in 2 different date?
What did I do?
I used the sum() to calculate time-range like this: df[~df['time'].dt.hour.between(10, 4)].sum()but it gives me sum as a whole of the df but not on the between time-range I have specified.
I also tried the resample but I cannot find a way to do it for time-specific
df['time'].dt.hour.between(10, 4) is always False because no number is larger than 10 and smaller than 4 at the same time. What you want is to mark between(4,21) and then negate that to get the other hours.
Here's what I would do:
# mark those between 4AM and 10PM
# data we want is where s==False, i.e. ~s
s = df['time'].dt.hour.between(4, 21)
# use s.cumsum() marks the consecutive False block
# on which we will take sum
blocks = s.cumsum()
# again we only care for ~s
(df[~s].groupby(blocks[~s], as_index=False) # we don't need the blocks as index
.agg({'time':'min', 'Open':'sum'}) # time : min -- select the beginning of blocks
) # Open : sum -- compute sum of Open
Output for random data:
time Open
0 2017-01-01 00:00:00 1.282701
1 2017-01-01 22:00:00 2.766324
2 2017-01-02 22:00:00 2.838216
3 2017-01-03 22:00:00 4.151461
4 2017-01-04 22:00:00 2.151626
5 2017-01-05 22:00:00 2.525190
6 2017-01-06 22:00:00 0.798234
an alternative (in my opinion more straightforward) approach that accomplishes the same thing..there's definitely ways to reduce the code but I am also relatively new to pandas
df.set_index(['time'],inplace=True) #make time the index col (not 100% necessary)
df2=pd.DataFrame(columns=['start_time','end_time','sum_Open']) #new df that stores your desired output + start and end times if you need them
df2['start_time']=df[df.index.hour == 22].index #gets/stores all start datetimes
df2['end_time']=df[df.index.hour == 4].index #gets/stores all end datetimes
for i,row in df2.iterrows():
df2.set_value(i,'sum_Open',df[(df.index >= row['start_time']) & (df.index <= row['end_time'])]['Open'].sum())
you'd have to add an if statement or something to handle the last day which ends at 11pm.
I am trying to group a data set of travel duration with 5 minutes interval, starting from 0 to inf. How may I do that?
My sample dataFrame looks like:
Duration
0 00:01:37
1 00:18:19
2 00:22:03
3 00:41:07
4 00:11:54
5 00:21:34
I have used this code: df.groupby([pd.Grouper(key='Duration', freq='5T')]).size()
And I have found following result:
Duration
00:01:37 1
00:06:37 0
00:11:37 1
00:16:37 2
00:21:37 1
00:26:37 0
00:31:37 0
00:36:37 1
00:41:37 0
Freq: 5T, dtype: int64
My expected result is:
Duration Counts
00:00:00 0
00:05:00 1
00:10:00 0
00:15:00 1
00:20:00 1
........ ...
My expectation is the index will start from 00:00:00 instead of 00:01:37.
Or, showing bins will also work for me, I mean:
Duration Counts
0-5 1
5-10 0
10-15 1
15-20 1
20-25 2
........ ...
I need your help please. Thank you.
First, you need to roud off your time to lower 5th minute. Then simply count it.
I suppose this is what you are looking for -
def round_to_5min(t):
""" This function rounds a timedelta timestamp to the nearest 5-min mark"""
t = datetime.datetime(1991,2,13, t.hour, t.minute - t.minute%5, 0)
return t
data['new_col'] = data.Duration.map(round_to_5min).dt.time
I have this huge dataset which has dates for several days and timestamps. The datetime format is in UNIX format. The datasets are logs of some login.
The code is supposed to group start and end time logs and provide log counts and unique id counts.
I am trying to get some stats like:
total log counts per hour & unique login ids per hour.
log count with choice of hours i.e. 24hrs, 12hrs, 6 hrs, 1 hr, etc and day of the week and such options.
I am able to split the data with start and end hours but I am not able to get the stats of counts of logs and unique ids.
Code:
from datetime import datetime,time
# This splits data from start to end time
start = time(8,0,0)
end = time(20,0,0)
with open('input', 'r') as infile, open('output','w') as outfile:
for row in infile:
col = row.split()
t1 = datetime.fromtimestamp(float(col[2])).time()
t2 = datetime.fromtimestamp(float(col[3])).time()
print (t1 >= start and t2 <= end)
Input data format: The data has no headers but the fields are given below. The number of days is not known in input.
UserID, StartTime, StopTime, GPS1, GPS2
00022d9064bc,1073260801,1073260803,819251,440006
00022d9064bc,1073260803,1073260810,819213,439954
00904b4557d3,1073260803,1073261920,817526,439458
00022de73863,1073260804,1073265410,817558,439525
00904b14b494,1073260804,1073262625,817558,439525
00022d1406df,1073260807,1073260809,820428,438735
00022d9064bc,1073260801,1073260803,819251,440006
00022dba8f51,1073260801,1073260803,819251,440006
00022de1c6c1,1073260801,1073260803,819251,440006
003065f30f37,1073260801,1073260803,819251,440006
00904b48a3b6,1073260801,1073260803,819251,440006
00904b83a0ea,1073260803,1073260810,819213,439954
00904b85d3cf,1073260803,1073261920,817526,439458
00904b14b494,1073260804,1073265410,817558,439525
00904b99499c,1073260804,1073262625,817558,439525
00904bb96e83,1073260804,1073265163,817558,439525
00904bf91b75,1073260804,1073263786,817558,439525
Expected Output: Example Output
StartTime, EndTime, Day, LogCount, UniqueIDCount
00:00:00, 01:00:00, Mon, 349, 30
StartTime and Endtime = Human readable format
Only to separate data with range of time is already achieved, but I am trying to write a round off time and calculate the counts of logs and uniqueids. Solution with Pandas is also welcome.
Edit One: I more details
StartTime --> EndTIime
1/5/2004, 5:30:01 --> 1/5/2004, 5:30:03
But that falls between 5:00:00 --> 6:00:00 . So this way count of all the logs in the time range is what I am trying to find. Similarly for others also like
5:00:00 --> 6:00:00 Hourly Count
00:00:00 --> 6:00:00 Every 6 hours
00:00:00 --> 12:00:00 Every 12 hours
5 Jan 2004, Mon --> count
6 Jan 2004, Tue --> Count
And so on Looking for a generic program where I can change the time/hours range as needed.
Unfortunately i couldn't find any elegant solution.
Here is my attempt:
fn = r'D:\temp\.data\dart_small.csv'
cols = ['UserID','StartTime','StopTime','GPS1','GPS2']
df = pd.read_csv(fn, header=None, names=cols)
df['m'] = df.StopTime + df.StartTime
df['d'] = df.StopTime - df.StartTime
# 'start' and 'end' for the reporting DF: `r`
# which will contain equal intervals (1 hour in this case)
start = pd.to_datetime(df.StartTime.min(), unit='s').date()
end = pd.to_datetime(df.StopTime.max(), unit='s').date() + pd.Timedelta(days=1)
# building reporting DF: `r`
freq = '1H' # 1 Hour frequency
idx = pd.date_range(start, end, freq=freq)
r = pd.DataFrame(index=idx)
r['start'] = (r.index - pd.datetime(1970,1,1)).total_seconds().astype(np.int64)
# 1 hour in seconds, minus one second (so that we will not count it twice)
interval = 60*60 - 1
r['LogCount'] = 0
r['UniqueIDCount'] = 0
for i, row in r.iterrows():
# intervals overlap test
# https://en.wikipedia.org/wiki/Interval_tree#Overlap_test
# i've slightly simplified the calculations of m and d
# by getting rid of division by 2,
# because it can be done eliminating common terms
u = df[np.abs(df.m - 2*row.start - interval) < df.d + interval].UserID
r.ix[i, ['LogCount', 'UniqueIDCount']] = [len(u), u.nunique()]
r['Day'] = pd.to_datetime(r.start, unit='s').dt.weekday_name.str[:3]
r['StartTime'] = pd.to_datetime(r.start, unit='s').dt.time
r['EndTime'] = pd.to_datetime(r.start + interval + 1, unit='s').dt.time
print(r[r.LogCount > 0])
PS the less periods you will have in the report DF - r, the faster it will count. So you may want to get rid of rows (times) if you know beforehand that those timeframes won't contain any data (for example during the weekends, holidays, etc.)
Result:
start LogCount UniqueIDCount Day StartTime EndTime
2004-01-05 00:00:00 1073260800 24 15 Mon 00:00:00 01:00:00
2004-01-05 01:00:00 1073264400 5 5 Mon 01:00:00 02:00:00
2004-01-05 02:00:00 1073268000 3 3 Mon 02:00:00 03:00:00
2004-01-05 03:00:00 1073271600 3 3 Mon 03:00:00 04:00:00
2004-01-05 04:00:00 1073275200 2 2 Mon 04:00:00 05:00:00
2004-01-06 12:00:00 1073390400 22 12 Tue 12:00:00 13:00:00
2004-01-06 13:00:00 1073394000 3 2 Tue 13:00:00 14:00:00
2004-01-06 14:00:00 1073397600 3 2 Tue 14:00:00 15:00:00
2004-01-06 15:00:00 1073401200 3 2 Tue 15:00:00 16:00:00
2004-01-10 16:00:00 1073750400 20 11 Sat 16:00:00 17:00:00
2004-01-14 23:00:00 1074121200 218 69 Wed 23:00:00 00:00:00
2004-01-15 00:00:00 1074124800 12 11 Thu 00:00:00 01:00:00
2004-01-15 01:00:00 1074128400 1 1 Thu 01:00:00 02:00:00